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Solve the non-linear inequality. Express the solution in interval notation. x^4 is greater than x^2

Mathematics
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\[x^4>x^2\]
I take the square root of both sides and get the plus or minus square root of x^2 is greater than x, then I take the square root of x^2 and x to get x is greater than the plus or minus square root of x
\[x ^{4}>x ^{2}\] \[x ^{4}-x ^{2}>0\] \[x ^{2}(x ^{2}-1)>0\] \[x ^{2}>0 or x ^{2}>1\] x>0 or x<-1 or x>1 then x<-1 and x>0 will be the interval...

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Other answers:

\(x^2(x^2-1)>0\) divide it by \(x^2\): \((x-1)(x+1)>0\) \(x\in(-\infty;-1)\bigcup(1;\infty)\)
why is it greater than 0
oh subtract the x^2?
yep//....
@znimon understood the process?
almost, what about after x^2(x^2-1) is greater than 0
\(x^2\) is always bigger than 0. It equals 0 when \(x=0\). But it doesn't satisfy the inequality.
took x^2 common in both the terms....
yep got that
nops.. while splitting the inequality x has to be >0
will you break it down after the x^2(x^2-1) is greater than 0
yep.... both the split terms are compared to the inequalities...
I understand how it works so I need to find all the solutions for x that satisfy the inequality
thanks now I'll try some on my own. Will you recommend a problem that has the same properties?
try this one... \[x ^{8}-x ^{2}<0\] remeber x^3 has different properties when compared with x^2 .... have a good time!!!!1
thanks
@znimon you're welcome :)))))))))))))))))
\[\ \ x^4\ge x^2\\ \ \ x^2[x^2]\ge x^2\\x^2\text{ is either 0 non-zero.}\\ \text{if }x^2\text{ is 0,}\\ \ \ 0\times0=0\\ \ \ 0 = 0\\ \text{therefore }x^2[x^2]\ge x^2\\ \text{otherwise,}\\ \ \ x^2\ge1\\ \ \ |x|\ge\sqrt1\]
is either 0 or non-zero **
solution is \[x <0, x <1] so (negative infinity, to 0 not including zero)?
for x^3-x^2 is less than 0

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